L01 – Fundamental theorem of arithmetic, the well-ordering property, division algorithm, ideal,greatest common divisor, and Bézout’s theorem

Discrete Mathematics2026年3月4日发布 芮和
568 130
5,669字
24–36 分钟

Elementary Number Theory

  • Divisibility
    primes, division algorithm, greatest common divisor, fundamemtal theorem of arithmetic
  • Congruences
    congruence, residue classes, Euler’s theorem
  • Application (Public-key encryption)
    RSA, modular arithmetics, square and multiply, Euclidean algorithm, prime number generation, CRT
  • Application (Key exchange)
    groups, subgroups, cyclic groups, Discrete logarithm, Diffie-Hellman key exchange

Divisibility

Notation:={0,1,2,};={0,±1,};={ba:a,b,a0}\mathbb{N}=\left\{0,1,2,\dots\right\};\,\mathbb{Z}=\left\{0,\pm 1,\dots\right\};\,\mathbb{Q}=\left\{\frac{b}{a}:\, a,b\in\mathbb{Z},\, a\neq 0\right\}(rational); ={a.b1b2,a,b1,b2,}\mathbb{R}=\left\{a.b_1b_2\dots,\,a,b_1,b_2,\dots\in\mathbb{Z}\right\}(real)

Definition: Let a{0}a\in\mathbb{Z}\setminus\left\{0\right\} and let bb\in\mathbb{Z}.

  • aadividesbb: there is an integer cc\in\mathbb{Z} such that b=acb=ac
  • aa is a divisor of bb; bb is a multiple of aa
  • a|ba\mid b: aa divides bb; aba\nmid b: aa does not divide bb
  • n{2,3,}n\in\left\{2,3,\dots\right\} is a prime if the only positive divisors of nn are 11 and nn
  • If n{2,3,}n\in\left\{2,3,\dots\right\} is not a prime, then nn is called a composite
  • Example: nn is composite iff a,b(1,n)\exist a,b\in\left(1,n\right)\cap\mathbb{Z} such that n=abn=ab

Theorem (Fundamental Theorem of Arithmetic):
Every integer n>1n>1 can be uniquely written as n=p1e1prern=p_1^{e_1}\dots p_r^{e_r}, where p1<<prp_1<\dots<p_r are primes and e1,,er1e1,\dots,e_r\geq 1.


FTA Proof

Existence: by mathematical induction on the integer nn

  • n=2n=2: 2=212=2^1 is a product of prime powers
  • Induction hypothesis: suppose there is an integer k>2k>2 such that the theorem is true for all integers nn such that 2n<k2\leq n<k
  • Prove the theorem is true for n=kn = k
    • n=kn=k is a prime
      • n=kn=k is a product of prime powers
    • n=kn=k is composite
      • There are integers n1,n2n_1,n_2 such that 1<n1,n2<n1<n_1,n_2<n and n=n1n2n=n_1n_2
      • By induction hypothesis, n1=p1α1prαrn_1=p_1^{\alpha_1}\dots p_r^{\alpha_r} and n2=q1β1qsβsn_2=q_1^{\beta_1}\dots q_s^{\beta_s}
        • p1,,pr,q1,,qsp_1,\dots,p_r,q_1,\dots,q_s are primes; α1,,αr,β1,,βs1\alpha_1,\dots,\alpha_r,\beta_1,\dots,\beta_s\geq 1
      • n=n1n2=p1α1prαrq1βqsβn=n_1n_2=p_1^{\alpha_1}\dots p_r^{\alpha_r}\cdot q_1^{\beta}\dots q_s^{\beta} is a product of prime powers

Division Algorithm

r=abx,{a=bq+r0r<br=a-bx,\,\begin{cases}a=bq+r\\0\leq r<b\end{cases}

The Well-Ordering Property: Every non-empty subset of \mathbb{N} has a least element.

THEOREM (Division Algorithm):
Let a,ba,b\in\mathbb{Z} and b>0b>0. Then there are unique q,rq,r\in\mathbb{Z} such that 0r<b0\leq r<b and a=bq+ra=bq+r.

  • Existence: Let S={abx:x and abx0}S=\left\{a-bx:\,x\in\mathbb{Z}\text{ and }a-bx\geq 0\right\}. Then
    • SS\neq\emptyset and SS\subseteq\mathbb{N}
      • SS has a least element, say r=abq0r=a-bq\geq 0
      • If rbr\geq b, then rb=ab(q+1)Sr-b=a-b(q+1)\in S and rb<rr-b<r.
        • The contradiction shows that 0r<b0\leq r<b.
  • Uniqueness: Suppose that q,r,0r<bq’,r’\in\mathbb{Z},\,0\leq r'<b and a=bq+ra=bq’+r’
    • Recall that a=bq+r,0r<ba=bq+r,\,0\leq r<b
      • Then b(qq)=rr(b,b)b\left(q-q’\right)=r’-r\in(-b,b)
        • It must be the case that q=qq=q’ and thus r=rr=r’

Ideal

{a=bq+r0r<b\begin{cases}a=bq’+r’\\0\leq r'<b\end{cases}

bq+r=bq+rbq’+r’=bq+r
b(qq)=rr(b,b)b\left(q’-q\right)=r-r’\in(-b,b)

DEFINITION: Let II \subseteq \mathbb{Z} be nonempty. II is called an ideal of \mathbb{Z} if:

  • a,bIa+bIa, b \in I \implies a + b \in I; and
  • aI,rraIa \in I, r \in \mathbb{Z} \implies ra \in I.
    • Example: d=0,±d,±2d,d\mathbb{Z} = {0, \pm d, \pm 2d, \dots} is an ideal of \mathbb{Z} for all dd \in \mathbb{Z}.

THEOREM: Let II be an ideal of \mathbb{Z}. Then d\exists d \in \mathbb{Z} such that I=dI = d\mathbb{Z}.

  • If I={0}I = \{0\}, then d=0d = 0.
  • Otherwise, let S={aI:a>0}S = \left\{a \in I :\,a > 0\right\}.
    • SS \subseteq \mathbb{N} and SS \neq \emptyset.
    • due to the well-ordering property, SS has a least element, say dSd \in S.
      • dId\mathbb{Z} \subseteq I:
        • dIdrId \in I \implies dr \in I for any rr \in \mathbb{Z}.
      • IdI \subseteq d\mathbb{Z}:
        • xI,x=dq+r,0r<d\forall x \in I,\,x = dq + r,\,0 \le r < d.
        • r=xdqI,0r<dr = x – dq \in I,\,0 \le r < d.
        • r=0r = 0 (otherwise, there is a contradiction)
        • x=dqdx = dq \in d\mathbb{Z}.

DEFINITION: Let I1,I2I_1, I_2 be ideals of \mathbb{Z}. Then the sum of I1I_1 and I2I_2 is defined as I1+I2={x+y:xI1,yI2}I_1 + I_2 = \left\{x + y :\, x \in I_1, y \in I_2\right\}

THEOREM: If I1,I2I_1, I_2 are ideals of \mathbb{Z}, then I1+I2I_1 + I_2 is an ideal of \mathbb{Z}.

  • a,bI1+I2,a+bI1+I2\forall a, b \in I_1 + I_2,\, a + b \in I_1 + I_2
    • x1,x2I1,y1,y2I2\exists x_1, x_2 \in I_1,\,y_1, y_2 \in I_2 such that a=x1+y1;b=x2+y2a = x_1 + y_1;\,b = x_2 + y_2
    • a+b=(x1+x2)+(y1+y2)I1+I2a + b = (x_1 + x_2) + (y_1 + y_2) \in I_1 + I_2
  • aI1+I2,r,raI1+I2\forall a \in I_1 + I_2,\,r \in \mathbb{Z},\,ra \in I_1 + I_2
    • xI1,yI2\exists x \in I_1,\,y \in I_2 such that a=x+ya = x + y
    • ra=(rx)+(ry)I1+I2ra = (rx) + (ry) \in I_1 + I_2

EXAMPLE:3+5=;4+6=23\mathbb{Z} + 5\mathbb{Z} = \mathbb{Z};\,4\mathbb{Z} + 6\mathbb{Z} = 2\mathbb{Z}

  • 3+53\mathbb{Z} + 5\mathbb{Z} \subseteq \mathbb{Z}: this is obvious
  • 3+5\mathbb{Z} \subseteq 3\mathbb{Z} + 5\mathbb{Z}:
    • For every n,n=3(2n)+5(n)3+5n \in \mathbb{Z},\,n = 3 \cdot (2n) + 5 \cdot (-n) \in 3\mathbb{Z} + 5\mathbb{Z}
a+b=gcd(a,b)a\mathbb{Z} + b\mathbb{Z} =\gcd(a,b)\cdot\mathbb{Z}

Greatest Common Divisor

DEFINITION: Let a,ba, b \in \mathbb{Z} and at least one of them is nonzero.

  • common divisor: an integer dd such that d|a,d|bd\mid a, d\mid b
  • greatest common divisorgcd(a,b)\gcd(a, b): the largest common divisor
    • relatively prime:gcd(a,b)=1\gcd(a, b) = 1

THEOREM: Let a,ba, b \in \mathbb{Z} and at least one of them is nonzero. Then a+b=gcd(a,b)a\mathbb{Z} + b\mathbb{Z} = \gcd(a, b)\mathbb{Z}.

  • a,b0a+b0{a, b} \neq {0} \implies a\mathbb{Z} + b\mathbb{Z} \neq {0}
  • There exists d0d \in \mathbb{Z} \setminus {0} such that a+b=da\mathbb{Z} + b\mathbb{Z} = d\mathbb{Z}. W.l.o.g., d>0d > 0.
    • dd is a common divisor of a,b:a1+b0da, b:\,a \cdot 1 + b \cdot 0 \in d\mathbb{Z}
    • dd is greatest: Suppose that dd’ is a common divisor of a,ba, b
      • d|a,d|bd’|a,\,d’|b
      • a+b=dd=as+bta\mathbb{Z} + b\mathbb{Z} = d\mathbb{Z} \implies d = as + bt for some integers s,ts, t
      • d|dd’\mid d and thus ddd’ \leq d

THEOREM (Bézout): There exist s,ts,t\in\mathbb{Z} such that gcd(a,b)=as+bt\gcd(a,b)=as+bt.

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13 条评论

  • 反骨精
    反骨精 读者

    这数学符号看着有点懵圈啊

    美国
    回复
  • 神经同步
    神经同步 读者

    贝祖定理这名字还挺有意思的

    英国
    回复
  • 夏树
    夏树 读者

    讲算术基本定理的证明还挺清楚的

    罗马尼亚
    回复
  • 山雾轻扬
    山雾轻扬 读者

    理想那部分有点绕,得再看几遍

    中国北京
    回复
  • RinBloom
    RinBloom 读者

    公钥加密应用这块挺酷的

    中国湖北
    回复
    • 梦回鲸歌
      梦回鲸歌 读者

      我也觉得挺有意思的

      中国陕西@ RinBloom
      回复
  • value20
    value20 读者

    欧几里得算法那部分有点没懂

    法国
    回复
  • 沉默的星尘
    沉默的星尘 读者

    整除定义那块儿有点抽象啊

    法国
    回复
  • 雾隐星瞳
    雾隐星瞳 读者

    理想的和最大公约数关系这块挺妙

    日本
    回复
    • 焰心之语
      焰心之语 读者

      这个联系确实很精妙

      中国湖南@ 雾隐星瞳
      回复
  • Noble Crane
    Noble Crane 读者

    贝祖定理证明蛮简洁的

    日本
    回复
  • 踏歌者
    踏歌者 读者

    数论课笔记整理得挺详细

    美国
    回复